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Loosely, a number is describable if it can be unambiguously defined by a finite string over a finite alphabet. Numbers such as $\frac{1}{3}$, $\sum_{n=0}^\infty \frac{1}{n!}$, and "the ratio of circumference to diameter of a circle" are all describable numbers. One can show that the set of all such numbers is countable.

Let $U$ be the set of indescribable real numbers. I "claim" $U$ is empty, and so every real number is describable. Suppose to the contrary. Place a well-order on $U$ (this can be done assuming the Axiom of Choice). The least element $u$ of $U$ admits the description "the least element of $U$ according to the specified well-order", contradicting the indescribability of $u$.

Obviously, something has gone awry here. I suspect it is some combination of my loose definition of describability and my self-referential "description" of $u$. My (slightly open-ended) question is "What's wrong here?".

P.S. There is a blog post addressing something rather like my question. As I understand it, the author's response is "the given description isn't really a description". If so, I'd enjoy some elaboration.

Austin Mohr
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  • I'm not sure if I had this discussion on the chat some long aeons ago, or was there a question about this some aeons ago on the main site. – Asaf Karagila Dec 02 '12 at 23:57
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    See Berry's Paradox ... http://en.wikipedia.org/wiki/Berry_paradox – GEdgar Dec 03 '12 at 00:02
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    Does the fact that it can be ordered lexicographically really mean that it is countable?? – Tengu Dec 03 '12 at 00:23
  • @Tengu: Where did lexicographically enter the picture? If you mean to imply that the fallacy here is that there are uncountable well-orders, note that the fallacy is rather that we use the well-ordering to generate more names, so paradoxically we can find a name for every number even if there are uncountably many of them. Because names are finite, this means there are only countably many of them so $U$ was countable to begin with. – Asaf Karagila Dec 03 '12 at 00:31
  • @AsafKaragila Thank you for your reply. I was just wondering why because, for instance, real number is in order but it's not countable. I just missed the finiteness of the names, though. Thank you anyway :) – Tengu Dec 03 '12 at 01:03
  • @AsafKaragila Tengu is referring to the line in my question: "By this definition, the set of describable numbers is countable, since the set can be lexicographically ordered." I had this phrase in my head from undergraduate days, but perhaps whoever taught it to me was jumping to conclusions (or else relying on extra assumptions that I've since forgotten). I suppose it is not correct, since one can order ${0,1}^\omega$ lexicographically. – Austin Mohr Dec 03 '12 at 01:03
  • @Austin: Note that the lexicographic order of ${0,1}^\omega$ is not a well-ordering. Indeed it is a dense order with minimum and maximum, and homeomorphic to the Cantor set. It is certainly not countable too. – Asaf Karagila Dec 03 '12 at 01:05
  • But ${0,1}^{<\omega}$ is countable, despite its lexicographical order (coupled with extension, of course) is not a well-ordering; we can prove this set is countable regardless. – Asaf Karagila Dec 03 '12 at 01:07
  • @AsafKaragila Right, but I had not yet gotten to the well-order. My previous erroneous claim was simply "If a set can be lexicographically ordered, then it is countable". – Austin Mohr Dec 03 '12 at 01:08
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    Ah, yes. This claim is indeed erroneous. But the set is countable after all, so I suppose no one was harmed in the process of writing this question. :-) – Asaf Karagila Dec 03 '12 at 01:09
  • You may be also interested in Löwenheim–Skolem theorem. Roughly it says that if a first-order logic theory has an infinite model, it has an infinite model of any given cardinality. In particular, it means that the theory of real numbers or ZFC have countable models. – Petr Dec 03 '12 at 11:24
  • https://www.smbc-comics.com/comic/interesting-4 – Jahan Claes Sep 15 '22 at 16:52

3 Answers3

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But you did not describe the well-ordering of $U$. You merely described the fact that it exists.

You extended your language, and this allowed you to describe another countable set of numbers. Alas repeating the argument until exhausting $U$ would require an uncountable number of iteration, and by that point the language is no longer finite.

Let me give a slightly more advanced analogy. One could argue that in $L$ (Godel's constructible universe) everything is definable including the well-ordering of $\mathbb R$. Recall that GCH holds in $L$ and for every countable $\alpha$ there are new real numbers added in $L_\alpha$, so the construction of the real number is not exhausted until $L_{\omega_1}$.

But in the construction of $L$ we allow parameters which were previously constructed to be used. This allows us to extend our language, so to speak, and by that to generate another countable set of real numbers. In order to exhaust the entire collection of real numbers we had to get to $L_{\omega_1}$, that is to say that we had to make uncountably many steps. So no countable iteration covered everything.

Asaf Karagila
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    In particular, the "extension" is by the use of some well ordering that does exist but which may not itself be describable in terms of the numbers that have already been described. – Carl Mummert Dec 03 '12 at 00:05
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    Carl, yes. Exactly. – Asaf Karagila Dec 03 '12 at 00:05
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    The construction of $L$ allows not only parameters that are previously-constructed, but arbitrary ordinals, which you don't even need to provide individual descriptions for -- they're given to you for free! (Or alternatively you could say that you get each of them successively from the same description: "wrap up all ordinals we've seen so far as a set, please"). – hmakholm left over Monica Dec 03 '12 at 00:31
  • @Carl, I wanted to add this but I was having second thoughts: The predicate "cannot be described" is itself undefinable internally, and we can only define it externally. Or at least it cannot be recognized internally (much like we may have a model which knows about the set of true formulas of itself, but does not know that this set is the set of true formulas). Am I correct? – Asaf Karagila Dec 03 '12 at 00:37
  • @Asaf: I would say it's not so much that the set is unrecognizable as it is that there's not a formula of set theory to even express the property we want to recognize. Compare the way we make the elementary diagram for a countable model: we begin with the atomic diagram and then repeatedly take the Turing jump of it. Each jump allows us to decide formulas with one more quantifier; the $\omega$th jump allows us to decide the whole elementary diagram. In the context of set theory the atomic diagram of a model is not even a set, internally to that model, so we can't even get off the ground. – Carl Mummert Dec 03 '12 at 01:19
  • @Carl: I was talking about the case where $\kappa$ is inaccessible, then the real number encoding the theory of $V_\kappa$ is an element of $V_\kappa$, but this is a model of ZFC which knows about this set, it just doesn't know that it is... the truth. – Asaf Karagila Dec 03 '12 at 01:22
  • Yes, that's true in some sense. There is a formula $\phi(D,S)$ in the language of set theory saying "$D$ is the elementary diagram of $S$ in the language of set theory". ZFC proves $(\forall S)(\exists D)\phi(D,S)$. Now $V_\kappa$ doesn't recognize that $V_\kappa$ is a set (nor $\kappa$), so that proof doesn't help $V_\kappa$ tell that its diagram is one of its members. But it's not that $V_\kappa$ can define "the elementary diagram of $V_\kappa$" but can't recognize it; the problem is that $V_\kappa$ cannot even find the words to express the property. This is in different than with $0\sharp$. – Carl Mummert Dec 03 '12 at 01:40
  • I don't get it. What's the conclusion? Is every number describable or not? – user13107 Dec 03 '12 at 09:00
  • @user13107: No, of course not. – Asaf Karagila Dec 03 '12 at 11:52
  • @Carl: Yes, I had this discussion a year or two ago with one of my teachers. The point of $0^#$ is that it is the machine which deciphers the truth of $L$, and the universe knows it. It's magical because we normally don't think about inner models as acting like set models, but suddenly the universe can know exactly what is true in $L$. Witchery! – Asaf Karagila Dec 03 '12 at 11:54
  • @Asaf: my memory is getting hazy at this point, but if I recall correctly a transitive model could contain the set $0^\sharp$, and it can always express its usual definition ("the theory of an uncountable set of order indiscernables for $L_a$ for some $a$") but unless there are sufficient large cardinals around the model may still not satisfy "$0^\sharp$ exists". Is that right? – Carl Mummert Dec 03 '12 at 12:30
  • @Carl: Hmmm. That's a good question, but I think that the answer is negative. Trim the universe at the least inaccessible, then $0^#$ still exists but there are no large cardinals present. In $L$, however, all the cardinals of the universe are still very large, so consistency-wise we didn't change too much. The common way of proving $0^#$ exists is by assuming Erdos/Ramsey/measurable cardinal exists and showing there are indiscernibles in $L$. Maybe that part confused you? – Asaf Karagila Dec 03 '12 at 12:40
  • @Asaf: Here is the sort of thing I was thinking about. Let $C$ be the set of natural numbers called "$0^\sharp$". Let $M$ be a countable transitive model of $ZF$ containing $C$. By a theorem of Barwise (I believe), $M$ has an end-extension to a (still transitive) model of $V=L$. Thus $C$ is contained in a transitive model of $V = L$. But that model can't satisfy the sentence "$0^\sharp$ exists" because that sentence implies $V \not = L$. I may have gone wrong somewhere, but I'm blind to it if I did. I don't think this is the same argument I heard long ago but it has the same kind of moral. – Carl Mummert Dec 03 '12 at 14:43
  • @Carl: I think that the end extension is not necessarily transitive. – Asaf Karagila Dec 03 '12 at 14:48
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Besides the issue with the axiom of choice possibly giving you an undefinable well ordering, there is still the usual issue that "unambiguously defined by a finite string over a finite alphabet" does not succeed in defining a set of real numbers. The actual set existence axiom in set theory only allows you to form a set of real numbers if the set is definable by a formula of set theory. The quoted phrase cannot, in fact, be expressed by a formula of set theory.

There is a very thorough answer by Joel David Hamkins located here, which explains the situation in more detail.

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Carl Mummert
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The problem is with your description “the least element of $U$ according to the specified well-order”. In fact you don’t have a specified well-order here: after all, you needed the Axiom of Choice even to show that such a well-order exists. Since the well-order is not specified, you are not describing a particular element of $U$; in other words your description isn’t really a description at all.

Actually, I suspect that with the appropriate amount of care your argument could be made into a proof that $U$ cannot be well-ordered without using the Axiom of Choice (since your argument shows there is no constructible well-ordering).

Robin Houston
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  • But what about a universe like $L$ where there is a canonical well-ordering of the universe? In such universe of sets $U$ can be well-ordered in a "relatively" definable way, which is sufficient to agree that we may include this canonical order in our language. – Asaf Karagila Dec 03 '12 at 00:11
  • @AsafKaragila Thank you. I see I was mistaken. – Robin Houston Dec 03 '12 at 09:01